Problem: The equation of hyperbola $H$ is $\dfrac {(y-4)^{2}}{36}-\dfrac {(x-2)^{2}}{25} = 1$. What are the asymptotes?
We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-4)^{2}}{36} = 1 + \dfrac {(x-2)^{2}}{25}$ Multiply both sides of the equation by $36$ $(y-4)^{2} = { 36 + \dfrac{ (x-2)^{2} \cdot 36 }{25}}$ Take the square root of both sides. $\sqrt{(y-4)^{2}} = \pm \sqrt { 36 + \dfrac{ (x-2)^{2} \cdot 36 }{25}}$ $ y - 4 = \pm \sqrt { 36 + \dfrac{ (x-2)^{2} \cdot 36 }{25}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 4 \approx \pm \sqrt {\dfrac{ (x-2)^{2} \cdot 36 }{25}}$ $y - 4 \approx \pm \left(\dfrac{6 \cdot (x - 2)}{5}\right)$ Add $4$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{6}{5}(x - 2)+ 4$